If $P$ is a point equidistant from all the vertices $A(-1, 3)$,$B(3, 5)$,and $C(5, 7)$ of a triangle $ABC$,then $PA=$

Let $PQR$ be an acute-angled triangle in which $PQ < QR$. From the vertex $Q$,draw the altitude $QQ_1$,the angle bisector $QQ_2$,and the median $QQ_3$,with $Q_1, Q_2, Q_3$ lying on $PR$. Then,

Assertion: If $(0, 3), (1, 1)$ and $(-1, 2)$ are the midpoints of the sides of a triangle,then the centroid of the original triangle is $(0, 2)$.
Reason: The centroid of a triangle and the centroid of the triangle formed by joining the midpoints of the sides of the original triangle are the same.

The distance (in units) between the circumcentre and the centroid of the triangle formed by the vertices $(1, 2)$,$(3, -1)$ and $(4, 0)$ is

The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$,where $a \ne 0$. Then for any $a$,the orthocentre of this triangle lies on the line:

If the midpoints of the sides of a triangle are $(-2, 3), (4, -3)$ and $(4, 5)$,then the centroid of the triangle is